02 · Intermediate

Piecewise & small-signal diode models

Turn the exponential diode law into controlled circuit approximations instead of folklore. Derive the ideal-diode complementarity model, constant-drop and piecewise-linear forward models, quantify their error against the exponential law, then linearize the exact I–V relation around a declared Q-point to obtain gd and rd. Keep large-signal and incremental variables separate, state the amplitude condition behind local linearization, and explain when junction/diffusion capacitances must be added for dynamic small-signal work.

180–230 minIntermediate
In this lesson
  • Derive the ideal-diode ON/OFF model as a limiting piecewise constitutive relation with mutually exclusive voltage/current conditions.
  • Use assumption-checking to select ON or OFF state and reject inconsistent guessed states.
  • Derive the constant-forward-drop model as an approximation to the exponential I–V law over a declared current range.
  • Build a piecewise-linear forward model vD≈Vγ+iDrf and interpret Vγ and rf as fitting parameters rather than universal physical constants.
  • Quantify approximation error by comparing model-predicted voltage/current with the exponential-law operating point.
  • Linearize iD(vD) around a bias Q-point using a first-order Taylor expansion.
  • Derive gd=(diD/dvD)|Q=(IDQ+IS)/(nVT) and rd=1/gd≈nVT/IDQ in strong forward bias.
  • Separate total variables vD=VDQ+vd and iD=IDQ+id from their small-signal increments.
  • State an explicit small-signal amplitude criterion from the neglected higher-order Taylor terms instead of saying only 'small enough'.
  • Explain why charge-storage/junction capacitances are absent from the static rd model and when they must be included in frequency-dependent diode models.
Starting question

The exponential diode law is physically meaningful but inconvenient for hand analysis. How can we simplify it without pretending the simplification is the device itself?

02.1Keep the circuit references fixed while changing models

Figure 2.1 — One bias network, several diode models

The one question is: which diode model replaces D without changing the iD and vD references?

+−VsRDGNDiD+−vD

The external network stays the same. Only the constitutive law assigned to D changes: exponential, ideal, constant-drop or piecewise-linear.

Every model in this chapter describes the same two-terminal voltage vD and current iD. We are not changing the circuit topology when we change diode model; we are replacing one constitutive relation with another.

Figure 2.2 — Approximations form a model hierarchy

The one question is: what is retained or discarded at each simplification?

exponentialpiecewise-linearconstant Vγideal

Simpler models solve faster, but their acceptable operating range must be stated more narrowly.

02.2Ideal diode as a complementarity model

With vD defined anode-to-cathode and iD entering the anode, the ideal diode is represented by two mutually exclusive branches:

ON: vD=0, iD≥0
OFF: iD=0, vD≤0

Equivalently,

iD≥0, vD≤0, iDvD=0

This is not the Shockley law with one parameter set. It is a limiting switch-like model: forward conduction has zero voltage, reverse blocking has zero current, and reverse breakdown is absent.

State-check method. Assume ON, replace D by a short, solve the linear circuit, then verify iD≥0. If inconsistent, reject it. Assume OFF, replace D by an open, solve, then verify vD≤0.

02.3Constant-drop model: useful only after choosing a range

A more useful hand model keeps a nonzero forward drop:

ON: vD≈Vγ, iD≥0
OFF: iD≈0, vD<Vγ

Vγ is a chosen approximation for a specified diode type, temperature and current range. It is not a universal “turn-on voltage.”

For the chapter-01 example (5 V, 1 kΩ, IS=1 pA, n=1, 300 K), the exponential solution was vD≈0.57415 V and iD≈4.42585 mA. Choosing Vγ=0.60 V predicts

iD,CD=(5−0.60)/1kΩ=4.40 mA

The current error is about −0.58%. Choosing 0.70 V would predict 4.30 mA, about −2.84% here. The “better” constant depends on the operating range being approximated.

02.4Piecewise-linear forward model adds a finite slope

Over a limited forward-current interval, approximate the exponential curve by a line:

vD≈Vγ+rfiD

Here Vγ is the line intercept and rf its fitted slope. Both depend on the interval and fitting rule.

Given two chosen points (I1,V1) and (I2,V2),

rf=(V2−V1)/(I2−I1)
Vγ=V1−rfI1

Inserted into the bias circuit,

Vs=R iD+Vγ+rfiD
iD≈(Vs−Vγ)/(R+rf)

This is a large-signal piecewise approximation. Its rf is not automatically the small-signal rd derived later.

02.5Model selection is an error-budget decision

A disciplined workflow is: choose the simplest model that can meet the required accuracy, solve, then compare against a more complete law or data at representative operating points.

ModelRetainsDiscards
Idealdirectionalityforward drop and slope
Constant dropdirectionality + typical dropforward slope/current dependence
Piecewise lineardirectionality + local large-signal slopecurvature outside fit interval
Exponentialstatic nonlinear curvaturemany high-current/reverse/dynamic effects

“More detailed” is not automatically “better” if its parameters are unknown or if the task does not require the extra detail.

02.6Small-signal model is a Taylor expansion around Q

Figure 2.3 — Around Q, the diode is locally an rd

The one question is: which quantities belong to the DC operating point and which belong to the small signal?

large signal / Q-pointDVDQ, IDQsmall signalrdvd, id→

First solve Q using the large-signal model. Only then, for a sufficiently small perturbation, replace the diode by rd in the incremental circuit.

Write the total voltage as a DC operating point plus an increment:

vD(t)=VDQ+vd(t)
iD(t)=IDQ+id(t)

Taylor-expand the nonlinear constitutive law around VDQ:

i(VDQ+vd)≈IDQ+[di/dv]Qvd

Define the incremental conductance and resistance

gd=[diD/dvD]Q
rd=1/gd
id≈gdvd=vd/rd

The small-signal circuit describes increments only. A resistor rd cannot replace the diode when computing its DC Q-point.

02.7Derive gd and rd from the exponential law

For the compact exponential model

iD=IS[exp(vD/(nVT))−1]

differentiate:

diD/dvD=ISexp(vD/(nVT))/(nVT)

At Q, because IDQ+IS=IS exp(VDQ/(nVT)),

gd=(IDQ+IS)/(nVT)
rd=nVT/(IDQ+IS)

In strong forward bias IDQ≫IS,

rd≈nVT/IDQ

Example: at 300 K, n=1 and IDQ=1 mA, rd≈25.85 Ω. At 10 mA it is ≈2.585 Ω. The incremental resistance therefore depends strongly on bias.

02.8How small is “small signal”?

For the forward exponential, the Taylor expansion around Q contains

id≈gdvd+[gd/(2nVT)]vd²+...

The magnitude ratio of the quadratic term to the linear term is approximately

|quadratic/linear|≈|vd|/(2nVT)

Thus, if we want that local second-order correction below a declared fraction ε, a sufficient first estimate is

|vd|≲2ε nVT

For ε=5%, n=1 and 300 K this gives about 2.6 mV peak. This is not a universal distortion specification—the surrounding circuit and waveform matter—but it turns “small enough” into a quantitative model check.

02.9Static rd is not the complete high-frequency small-signal diode

The rd model comes only from the slope of the static I–V relation. A real p–n junction also stores charge.

Reverse/depletion bias produces junction (depletion) capacitance; forward injected minority charge produces diffusion/storage capacitance. A frequency-dependent small-signal model therefore adds capacitances and, depending on required accuracy, parasitic series resistance and packaging elements.

Yd(jω)≈gd+jω(Cj+Cd)

This is a preview, not a universal diode RF model: Cj and Cd depend on bias and device physics. The pure rd approximation is appropriate only when the relevant reactive currents are negligible over the frequency range of interest.

02.10Mastery exercises

  1. In the bias network, assume an ideal diode ON. What consistency condition must the solved current satisfy?
  2. Assume it OFF. What sign must the solved vD satisfy?
  3. For Vs=5 V,R=1 kΩ,Vγ=0.65 V, find the constant-drop forward current.
  4. Fit a line through (1 mA,0.55 V) and (5 mA,0.59 V). Find rf and Vγ.
  5. Use that affine model with Vs=5 V,R=1 kΩ to estimate current.
  6. At T=300 K,n=1,IDQ=2 mA, find rd.
  7. At n=1 and 300 K, what peak vd makes the Taylor quadratic term about 10% of the linear term?
  8. Explain why using rd to compute the DC bias point is circular.
Solutions

1. iD≥0.

2. vD≤0.

3. (5−0.65)/1kΩ=4.35 mA.

4. rf=(0.59−0.55)/(4 mA)=10 Ω; Vγ=0.55−10Ω·1mA=0.540 V.

5. iD≈(5−0.540)/(1010 Ω)≈4.416 mA.

6. rd≈25.85 mV/2 mA≈12.93 Ω.

7. |vd|≈2(0.10)VT≈5.17 mV.

8. rd depends on IDQ, which must first be obtained from the nonlinear or an appropriate large-signal model.

Chapter 03 will use these model-selection rules to derive rectifier operation over full input cycles, including conduction intervals, load effects, ripple assumptions and the difference between ideal and finite-drop predictions.